Find minimum of f(x) = x^2
Know that f'(x) = 2*x = 0 so
then 0/2 = 0
finally the answer is minimum of x^2 is 0.
Next homework help data.
so y'(x) = 2*x + 7 -2*3*x^-3.
set it equal to zero
0 = 2*x+7-6*x^-3
0*x^3=2*x^5+7*x^3-6
0=2*x^5+7*x^3-6.
solve for x. see Maple work